20070119, 22:28  #1 
Oct 2006
2^{2}·5·13 Posts 
Prime creator through sequence
While going through the internet, I came across a sequence of rules that (I think) guaranteed a prime number.
It was called "the repeated factorization of concatenated prime factors." The steps are : Choose a composite number Factor the number Put the factors in ascending (smallest>largest) order without spaces. This is your new number If the result is prime, you're done. If the result is composite, go back to step 2. I tested it out on some very small numbers, and then on a 6digit one (30+ iterations). Currently I am working on a 18digit composite, which on iteration 60, has revealed a C105, factored down from a ~C110. I'm using Msieve for this, and it appears to be the fastest for these numbers (below 110 digits). Anyone else interested? Roger Last fiddled with by roger on 20070119 at 22:57 
20070120, 03:07  #2 
"Jason Goatcher"
Mar 2005
5×701 Posts 
that's interesting. I'll give it a try with alpertron's app.

20070120, 03:18  #3 
"Jason Goatcher"
Mar 2005
5·701 Posts 
I got sick of editing after about 5 minutes. Maybe someone could write a program, possibly using Alpertron's app, that could test if this method is better than random chance.

20070120, 08:05  #4 
Jun 2003
12FD_{16} Posts 
http://www.mersenneforum.org/showthread.php?t=3238 for more info

20070120, 17:05  #5 
Jan 2005
Transdniestr
767_{8} Posts 
So basically, the only counterexample would be a number that ultimately entered into a loop (kind of like an aliquot sequence).
I wrote a little function in PARI to do this. Just to get the bowl rolling: mcat(a)={ print("Processing: ",a); while(!ispseudoprime(a), sa = 1; fs=factor(a); for(j=1,length(fs[,1]), for(k=1,fs[j,2], if (k==1,multp = 10^floor(1+log(fs[j,1])/log(10) ) ); if(sa==1, sa=fs[j,1] , sa = sa* multp + fs[j,1] ) ) ); print (" ",sa," vs. ",fs); a=sa ) } 
20070120, 19:35  #6 
"William"
May 2003
New Haven
3×787 Posts 
Isn't this the "Home Prime" problem?
http://mathworld.wolfram.com/HomePrime.html (Oh  I see axn1 made the same point through his link) Last fiddled with by wblipp on 20070120 at 19:37 Reason: Acknowledge axn1's prior point 
20070121, 04:00  #7 
Oct 2006
2^{2}×5×13 Posts 
Yeah, this is a HomePrime sequence.
The 'official' websites are Maintained by Patrick De Geest 1. mailto:pdg@worldofnumbers.com 2. mailto:Patrick.DeGeest@skynet.be Website 1 : http://www.worldofnumbers.com/index.html Website 2 : http://users.skynet.be/worldofnumbers/ (mirrorsite) To GrandpaScorpion: how is your code implemented so that PARI can use it? I pasted the code, and tried changing the sa value, but all that happened was that I could change the code. I have almost no experience with coding, so if possible, laymans terms are greatly appreciated! Thanks, Roger 
20070122, 17:47  #8  
∂^{2}ω=0
Sep 2002
República de California
2·3^{3}·5·43 Posts 
Quote:
How is this different in fundamental nature from the simpler: 1) Start with an odd number >= 1; 2) Add 2. 3) If the result is prime, you're done. 4) If the result is composite, go back to step 2. In fact, the add2 primegenerating "algorithm" is exponentially superior in terms of runtime, since it requires no factorization, just a primality test. Last fiddled with by ewmayer on 20070122 at 17:48 

20070126, 12:12  #9 
Feb 2006
3·17 Posts 
This code is too advanced for me  i was thinking more like:
1) Start with any integer >= 1 2) Add 1 3) If the result is prime, you're done. 4) Go back to step 2. Eivind Last fiddled with by Eivind on 20070126 at 12:16 
20070128, 01:42  #10 
"Jason Goatcher"
Mar 2005
3505_{10} Posts 
Not because I think it's a good algorithm, but simply because I think it would be fun, I'm going to check the answers on the home prime effort for 49 after the last major factorization, and maybe try to continue the sequence.

20070128, 12:37  #11  
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across
10100101100001_{2} Posts 
Quote:
I don't at present understand how you reached your conclusion and would like to be educated. Alex Kruppa and I have spent quite a bit of effort on the HP(49) sequence and it has yet to terminate. Paul 

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